Graph log base 4 of x-1

$\log_{4} (x - 1)$

Find the asymptotes.

Set the argument of the logarithm equal to zero.

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The vertical asymptote occurs at $x = 1$ .

Vertical Asymptote: $x = 1$

Find the point at $x = 2$ .

Replace the variable $x$ with $2$ in the expression.

$f (2) = \log_{4} ((2) - 1)$

Simplify the result.

Subtract $1$ from $2$ .

$f (2) = \log_{4} (1)$

Logarithm base $4$ of $1$ is $0$ .

$f (2) = 0$

The final answer is $0$ .

$0$

Convert $0$ to decimal.

$y = 0$

Find the point at $x = 5$ .

Replace the variable $x$ with $5$ in the expression.

$f (5) = \log_{4} ((5) - 1)$

Simplify the result.

Subtract $1$ from $5$ .

$f (5) = \log_{4} (4)$

Logarithm base $4$ of $4$ is $1$ .

$f (5) = 1$

The final answer is $1$ .

$1$

Convert $1$ to decimal.

$y = 1$

Find the point at $x = 3$ .

Replace the variable $x$ with $3$ in the expression.

$f (3) = \log_{4} ((3) - 1)$

Simplify the result.

Subtract $1$ from $3$ .

$f (3) = \log_{4} (2)$

Logarithm base $4$ of $2$ is $\frac{1}{2}$ .

Rewrite as an equation.

$\log_{4} (2) = x$

Rewrite $\log_{4} (2) = x$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b$ does not equal $1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$4^{x} = 2$

Create expressions in the equation that all have equal bases.

${(2^{2})}^{x} = 2^{1}$

Rewrite ${(2^{2})}^{x}$ as $2^{2 x}$ .

$2^{2 x} = 2^{1}$

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$2 x = 1$

Solve for $x$ .

$x = \frac{1}{2}$

The variable $x$ is equal to $\frac{1}{2}$ .

$f (3) = \frac{1}{2}$

The final answer is $\frac{1}{2}$ .

$\frac{1}{2}$

Convert $\frac{1}{2}$ to decimal.

$y = 0.5$

The log function can be graphed using the vertical asymptote at $x = 1$ and the points $(2, 0), (5, 1), (3, 0.5)$ .

Vertical Asymptote: $x = 1$

$\begin{array}{c} x & y \\ 2 & 0 \\ 3 & 0.5 \\ 5 & 1 \end{array}$

Graph log base 4 of x-1

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